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3 years ago

If I put a different sticker on the 30th, 50th, and 60th pages will any of the first 600 pages have all 3 stickers?

Mathematics

1 answer:

icang [17]3 years ago

8 0

Answer:

so the which page that will have the 3 sticker, we must solve the least common multiple of 30, 50, 60. A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24, ....

The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.

so the least common multiple of 30, 50 and 60 is 300. so the page that will have 3 stickers is 300th page

Step-by-step explanation:

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In a seventh grade class, the ratio of boys to girls is 3/9. there are 28 students in the class. Which equation show how to find

Shkiper50 [21]

Answer:

Hey there! ok , so your equation would be 28 / 3

Step-by-step explanation:

The equation would be 28/3 because there are 28 students in the class and the ratio of boys to girls is 3-9 meaning that it is 1/3, which is the same as dividing by 3, so the equation would be 28/3

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3 years ago

Analyze the diagram below and complete the instructions that follow. 550 450

svetlana [45]

Answer:

130 degrees

Step-by-step explanation:

let the angle be x

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4 years ago

How many years is 694.44 days?

antoniya [11.8K]

Answer:

1.9 years

Step-by-step explanation:

There are 365 days in a year

694.44 / 365 = 1.9

Hope this helps!

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3 years ago

Read 2 more answers

Help please!!! I dont understand these questions currently attaching photos dont delete

Katyanochek1 [597]

Answer:

b/a
16a²b²
n¹⁰/(16m⁶)
y⁸/x¹⁰
m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

$a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}$

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

$\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}$

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

$\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2$

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

$\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}$

4. This works the same way the previous problem does.

$\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}$

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

$\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}$

3 0

3 years ago

What is the length of the hypotenuse, c? Round your answer to the nearest hundredth.

Afina-wow [57]

Answer:

Step-by-step explanation:

By using the Pythagorean theorem:

b^2+3^2=5^2
b^2+9=25

b=sqrt (25-9)

b=sqrt(16)

b=4

=> 4^2+7^2=c^2
16+49=c^2
65=c^2
=>c=sqrt(65)

c = (approximately) 8

5 0

2 years ago